Problem: A secant line intersects the curve $y=x^2-3x$ at two points with $x$ -coordinates $5$ and $t$, where $t\neq 5$. What is the slope of the secant line in terms of $t$ ? Your answer must be fully expanded and simplified.
Explanation: We are given that the secant line intersects the curve at $x=5$ and $x=t$. Since these points are on the curve $y=x^2-3x$, we know that their $y$ -values are $y=10$ and $y=t^2-3t$, correspondingly. To summarize this part, we know that the secant line passes through the points $(5,10)$ and $(t, t^2-3t)$. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{t^2-3t-10}{t-5} \end{aligned}$ We can now simplify the expression we obtained. $\begin{aligned} \dfrac{t^2-3t-10}{t-5}&=\dfrac{(t-5)(t+2)}{t-5} \\\\ &=t+2\text{, for }t\neq 5 \end{aligned}$ Since we are given that $t\neq 5$, we can conclude that the slope of the secant line is $t+2$.